estimate a integral with parameter

Consider the following integral: $$J(u)=\int_{0}^{u^2}\int_{0}^{u}\frac{1}{x^3+y+3x^2+5x+3}dxdy$$ My question: I want to separate it into two parts: $$J(u)=u^aI(u)$$ where $$a\geq 0$$ and $$I(u)$$ is "smaller" than polynomials, that is, $$I(u)$$ satisfies the followings:

(1) $$\liminf\limits_{u\rightarrow+\infty}I(u)>0$$. (2) for any $$p>0$$, $$\lim\limits_{u\rightarrow+\infty}\frac{I(u)}{u^p}=0$$

For example, $$I(u)=\log(u)$$, or $$\arctan u$$ etc.

My attempt: I first guess $$a=0$$, that is, $$J(u)$$ itself satisfies $$I(u)$$. But I can't show that property (2) above. Then I guess $$a=1$$, that is, $$J(u)=uI(u)$$, I can show that property (2) by change the variables $$(\frac{x}{u},\frac{y}{u^2})=(w,s)$$ and using Dominated convergence theorem. But this case I can't show the property (1)... And now I use computer to calculate, I guess $$a$$ may be between $$(0,1)$$.

Some tips:

(i) if you change the variables, $$(\frac{x}{u},\frac{y}{u^2})=(w,s)$$ may be better. this is called dilation and it is a kind of "homogeneous"

(ii) If I change $$x^3$$ into $$xy$$ in $$J(u)$$, I can show that actually $$a=0$$, by Dominated convergence theorem.

Background: Given a dilation in $$R^n$$: $$D_t(x)=(t^{c_1}x_1,...,t^{c_n}x_n)$$ then consider $$P(x,r)=r^nf_n(x)+...+r^Qf_Q(x)$$, where $$Q=m_1+...+m_n$$, $$f_k(x)$$ satisfies $$f_k(D_t(x))=t^{Q-k}f_k(x)$$, and $$f(x)>0$$ are the combinations of positive monomials ($$x_i>0$$). The origin type of the denominator of the $$J(u)$$ is $$x^3r^3+(y+3x^2)r^4+5xr^5+3r^6$$. And the dilation is $$D_t(x)=(t^{1}x,t^2y,t^{3}z)$$. I just take $$1/r=u$$ and do some simplification.

• One direct simplification is to change the order of integration, to get $$\begin{split} J(u) &= \int_0^{u^2}\int_0^u\frac{dxdy}{x^3+y+3x^2+5x+3} \\ &= \int_0^u \left[ \int_0^{u^2}\frac{dy}{x^3+y+3x^2+5x+3} \right] dx \\ &= \int_0^u \ln \left(\frac{x^3+3x^2+5x+3} {u^2+x^3+3x^2+5x+3} \right) dx \\ &= \int_0^u \ln \left(1 - \frac{u^2} {u^2+x^3+3x^2+5x+3} \right) dx \\ \end{split}$$ Jun 14, 2020 at 4:49
• but, maybe you calculate wrong? the integral should be positive but yours is nagetive... and if you calculate in the right way you will find that's not a good way to due with the integral
– Houa
Jun 14, 2020 at 5:05
• We have $$J(u) = \int_0^u \ln \left( 1 + \frac {u^2} {P(x)} \right) dx, \\ J'(u) = 2 u \int_0^u \frac {dx} {P(x) + u^2} + \ln \left( 1 + \frac {u^2} {P(u)} \right).$$ The integral in $J'(u)$ (which is similar to this integral) can be brute-forced by writing out an antiderivative in terms of a sum over the roots of $P(x) + u^2$ and taking an asymptotic expansion of the roots for large $u$. This gives $$J'(u) \sim \frac {4 \pi} {3 \sqrt 3 \hspace {1.5px} u^{1/3}},$$ therefore $$J(u) \sim \frac {2 \pi u^{2/3}} {\sqrt 3}$$ by l'Hopital's rule. Jul 26, 2020 at 20:43
• @Maxim Good idea by l'Hopital's rule! By the way, do you have any reference to due with such integrals? This is just a concrete example. The more general type is like the description in the "background" in the body above.($P(x,r)=....$)
– Houa
Jul 27, 2020 at 1:00